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%%文档的题目、作者与日期
\author{王立庆（2019级数学与应用数学1班）}
\title{数量金融实验 - 第2章课堂练习}
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%\date{2022 年 9 月 8 日}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item %1
有关期权定价的二叉树方法，下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  这个方法是在1970年代末由 Sharpe, Cox, Ross 和 Rubinstein 等人提出来的。
\item[B.]  这个方法假定在给定的时间间隔内，股票的价格运动都有向上、向下和保持不变这三种可能。
\item[C.]  这个方法假定股票价格的每次向上或向下运动的概率和幅度都不变。
\item[D.]  这个方法模拟了在期权的有效期内股票价格的所有可能的发展路径。
\end{enumerate}

{\color{red}
解答：B. 这个方法假定在给定的时间间隔内，股票的价格运动只有向上和向下两个方向。

}

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\item %2
假设某个股票的当前价格是 50 美元，一年后，它的价值可能是 60 美元或者 45 美元。如果一年后相应的期权的价格为 $V_1^u=10$ 美元和 $V_1^d=0$ 美元。设一年期无风险利率为 5\%. 求 $t=0$ 时该期权的价格。
\begin{enumerate}
\item[A.]  2.80.
\item[B.]  3.80.
\item[C.]  4.80.
\item[D.]  5.80.
\end{enumerate}

{\color{red}
解答：C. 先计算股权的变化与股价的变化的比例，即德尔塔量，$$\Delta = \frac{10-0}{60-45}=\frac{2}{3}.$$
再由单期二叉树期权的定价公式，可得 
\begin{eqnarray*}
V_0 &=& \Delta S_0 + e^{-rT}(V_T^u-\Delta S_T^u) \\ 
&=& \frac{2}{3}\times 50 + e^{-0.05}\times \left(10-\frac{2}{3}\times 60 \right) = 4.7965.
\end{eqnarray*}

}

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\item %3
在期权定价的单期二叉树模型中，使用复制方法来导出期权的定价公式。下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  设无风险资产是债券，在 $t=0$ 时的单价是1. 设无风险利率是 $r$. 
\item[B.]  设投资组合 $\Phi_0=\alpha S_0+\beta$ 是期权 $V_0$ 的复制。
\item[C.]  根据复制的含义，在到期日 $t=T$ 时，这个投资组合的价值 $\Phi_T$ 与期权的价值 $V_T$ 相同。
\item[D.]  由二叉树模型假设，在到期日 $t=T$ 时，股票的价格有两种可能，分别记为 $S_T^u$ 和 $S_T^d$.  
\item[E.]  由二叉树模型假设，在到期日 $t=T$ 时，期权的价格也有两种可能，分别记为 $V_T^u$ 和 $V_T^d$. 
\item[F.]  由 $\Phi_T=V_T$, 可以将这个投资组合的系数 $\alpha$ 和 $\beta$ 分别写成 $S_T^u, S_T^d$ 和 $V_T^u, V_T^d$ 的表达式。
\item[G.]  记 $q=\frac{e^{rT}S_0-S_T^d}{S_T^u-S_T^d}$, 则在任何市场总有 $0<q<1$, 称为风险中性概率。
\item[H.]  最后得出在 $t=0$ 时的期权价格为 $V_0=\Phi_0=e^{-rT}[qV_T^u + (1-q)V_T^d]$. 这个公式可以理解成到期日的期权价格在风险中性概率下的数学期望，在 $t=0$ 时刻的贴现值。
\end{enumerate} 

{\color{red}
解答：G. 因为这里的上下文中，对无风险利率和到期日的股价的可能取值没有限制条件， 所以选项 G 里定义的这个表达式 $q$ 的取值范围不好确定。在无套利市场，这个范围是成立的。

}

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\item %4
假设某个股票的当前价格是 $S_0=50$ 美元，一年后，它的价值可能是 $S_1^u=60$ 美元或者 $S_1^d=45$ 美元。如果一年后相应的期权的价格为 $V_1^u=10$ 美元和 $V_1^d=0$ 美元。设一年期无风险利率为 $r=5\%$. 求这一年中这个股票上涨的风险中性概率 $q$. 
\begin{enumerate}
\item[A.]  0.5000.
\item[B.]  0.5042.
\item[C.]  0.6666.
\item[D.]  0.3333.
\end{enumerate}

{\color{red}
解答：B. 直接代入风险中性概率的公式，可得 
$$q=\frac{e^{rT}S_0-S_T^d}{S_T^u-S_T^d}= \frac{e^{0.05\times 1}\times 50 - 45}{60-45} = 0.5042.$$
由此可得初始时刻的期权价格，即到期日的期权价格的数学期望的贴现，为
$$V_0 = e^{-rT}[qV_1^u+(1-q)V_1^d] = e^{-0.05}[0.5042\times 10 + 0.4958*0] = 4.7961.$$
}

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\item %5
设一个离散时间随机过程 $\{X_k\mid k\ge 0\}$ 是一个鞅。记 $\mathcal{F}_k=\sigma(X_0,X_1,\cdots,X_k)$ 是由随机变量 $X_0,X_1,\cdots,X_k$ 生成的 $\sigma$-域。则在下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  对每个 $k\ge 0$, 一定有 $\mathbb{E}[|X_k|]<\infty$, 即每个随机变量都是绝对可积的。
\item[B.]  对每个 $k\ge 0$, 一定有 $\mathbb{E}[X_{k+1} \mid \mathcal{F}_k ] = X_k$, 这是鞅的最重要的特征。 
\item[C.]  由随机变量 $X_k$ 生成的 $\sigma$-域 $\sigma(X_k)$ 是包含事件族 $\{\{X_k\le t\} \mid t\in\mathbb{R}\}$ 的最小 $\sigma$-域。
\item[D.]  若 $\mathcal{F}$ 和  $\mathcal{G}$ 是集合 $\Omega$ 上的两个 $\sigma$-域，则  $\mathcal{F} \cup \mathcal{G}$ 也是集合 $\Omega$ 上的一个 $\sigma$-域。
\end{enumerate}

{\color{red}
解答：D. 设 $A\in\mathcal{F}$, $B\in\mathcal{G}$, 则 $A\cup B$ 不一定落在 $\mathcal{F} \cup \mathcal{G}$ 中。

}

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\item %6
设离散型二维随机变量 $(X,Y)$ 的分布列如下，

\begin{center}
\begin{tabular}{|c|c|c|c|c|} \hline 
联合概率 & $Y=0$ & $Y=1$ & $Y=2$ & 边际概率 \\  \hline 
$X=0$ & 0.1 & 0.2 & 0.3 &  0.6 \\ \hline 
$X=1$ & 0.2 & 0.1 & 0.1 &  0.4 \\ \hline 
边际概率 & 0.3 & 0.3 & 0.4 & 1 \\ \hline 
\end{tabular}
\end{center}
则条件期望 $\mathbb{E}[X\mid Y=1]$ 等于多少？

\begin{enumerate}
\item[A.]  1/2.
\item[B.]  1/3.
\item[C.]  1/4.
\item[D.]  1/6.
\end{enumerate}

{\color{red}
解答：B. 根据离散型随机变量的条件期望的计算公式，可得
\begin{eqnarray*}
\mathbb{E}[X\mid Y=1] = \sum\limits_{k=0,1} k\mathbb{P}[X=k\mid Y=1] 
= \mathbb{P}[X=1\mid Y=1] 
= \frac{\mathbb{P}[X=1, Y=1]}{\mathbb{P}[Y=1]} 
= \frac{0.1}{0.3} = \frac{1}{3}. 
\end{eqnarray*}

}

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\item %7
设在投资周期为 $[0,T=1]$ 的单期二叉树模型中，风险资产的价格下跌或上涨的幅度分别为 $d=0.9$ 和 $u=1.1$. 设无风险利率为 $r=0.1$. 下述说法中，正确的是哪个？

\begin{enumerate}
\item[A.]  因为 $d<e^{rT}<u$, 所以不存在套利。
\item[B.]  因为 $u<e^{rT}<d$, 所以不存在套利。
\item[C.]  因为 $u<e^{rT}$, 所以存在套利。
\item[D.]  因为 $d<e^{rT}$, 所以存在套利。
\end{enumerate}

{\color{red}
解答：C. 计算可得 $e^{rT}=e^{0.1}=1.1052>u$, 所以无风险收益超过了风险资产的上涨幅度，所以存在套利。

}


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\item %8
假设某个股票的当前价格是60美元，到期时间为6个月，股票价格每3个月上升或下降的幅度为 10\%, 无风险利率为 5\%. 求一个敲定价格为58美元的欧式看涨期权的价格。
\begin{enumerate}
\item[A.]  3.1838.
\item[B.]  4.1838.
\item[C.]  5.1838.
\item[D.]  6.1838. 
\end{enumerate}

{\color{red}
解答：C. 这里 $S_0=60, u=1.1, d=0.9, X=58, r=0.05, T=2$, 首先画出股票价格的二叉树图形，

\begin{center}
\tikz {
\node (dd) at (0,0) {$S_0=60$};  
\node (e1) at (3,0.5) {$S_0u=66$};  
\node (e2) at (3,-0.5) {$S_0d=54$};  
\node (f1) at (6,1) {$S_0u^2=72.6$};  
\node (f2) at (6,0) {$S_0ud=59.4$};  
\node (f3) at (6,-1) {$S_0d^2=48.6$};  

\graph { (dd) -- (e1) };
\graph { (dd) -- (e2) };
\graph { (e1) -- (f1) };
\graph { (e1) -- (f2) };
\graph { (e2) -- (f2) };
\graph { (e2) -- (f3) };
}
\end{center}

然后画出期权的二叉树图形，并写出6个月时（到期日）的三个节点的看涨期权的价格，

\begin{center}
\tikz {
\node (aa) at (0,0) {$c_0$};  
\node (b1) at (3,0.5) {$c_u$};  
\node (b2) at (3,-0.5) {$c_d$};  
\node (c1) at (6,1) {$c_{uu}=14.6$};  
\node (c2) at (6,0) {$c_{ud}=1.4$};  
\node (c3) at (6,-1) {$c_{dd}=0$};  

\graph { (aa) -- (b1) };
\graph { (aa) -- (b2) };
\graph { (b1) -- (c1) };
\graph { (b1) -- (c2) };
\graph { (b2) -- (c2) };
\graph { (b2) -- (c3) };
}
\end{center}

设 $t=1/4=0.25$ 表示3个月为一期，风险中性概率为 
$$q=\frac{e^{rt}-d}{u-d} = \frac{e^{0.05\times 0.25}-0.9}{1.1-0.9} = 0.5629.$$

根据风险中性测度 $Q=(q,1-q)=(0.5629,0.4371)$, 可得3个月时的两个节点的看涨期权的价格，分别为
\begin{eqnarray*}
c_u &=& e^{-rt} [qc_{uu}+(1-q)c_{ud}] = e^{-0.05\times 0.25}\times ( 0.5629\times 14.6 + 0.4371\times 1.4 ) = 8.7206, \\ 
c_d &=& e^{-rt} [qc_{ud}+(1-q)c_{dd}] = e^{-0.05\times 0.25}\times ( 0.5629\times 1.4 + 0.4371\times 0 ) = 0.7783. 
\end{eqnarray*}

最后得到 $t=0$ 时的看涨期权的价格，
\begin{eqnarray*}
c_0 = e^{-rt} [qc_{u}+(1-q)c_{d}] = e^{-0.05\times 0.25}\times ( 0.5629\times 8.7206 + 0.4371\times 0.7783 ) = 5.1838. 
\end{eqnarray*}

%注意到这个题目与书上例子2.2 的唯一区别是到期日不同，书上为9个月，最后的欧式看涨期权的价格为 6.68美元，比本题到期日为6个月的期权价格稍多一些。

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\end{enumerate}

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